Problem: Simplify the following expression: $\dfrac{36q^5}{9q^2}$ You can assume $q \neq 0$.
Explanation: $ \dfrac{36q^5}{9q^2} = \dfrac{36}{9} \cdot \dfrac{q^5}{q^2} $ To simplify $\frac{36}{9}$ , find the greatest common factor (GCD) of $36$ and $9$ $36 = 2 \cdot 2 \cdot 3 \cdot 3$ $9 = 3 \cdot 3$ $ \mbox{GCD}(36, 9) = 3 \cdot 3 = 9 $ $ \dfrac{36}{9} \cdot \dfrac{q^5}{q^2} = \dfrac{9 \cdot 4}{9 \cdot 1} \cdot \dfrac{q^5}{q^2} $ $\phantom{ \dfrac{36}{9} \cdot \dfrac{5}{2}} = 4 \cdot \dfrac{q^5}{q^2} $ $ \dfrac{q^5}{q^2} = \dfrac{q \cdot q \cdot q \cdot q \cdot q}{q \cdot q} = q^3 $ $ 4 \cdot q^3 = 4q^3 $